Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
double dp[3000][3000],s[3000];
int vis[3000][3000],n;
double fun(int h=0,int t=0)
{
    if(t>n/2)return 0;
    if(t+h==n)return 1;
    if(vis[h][t])return dp[h][t];
    vis[h][t]=1;
    return dp[h][t]= (fun(h+1,t)*s[h+t]+fun(h,t+1)*(1.0-s[h+t]));


}

int main()
{
    cin>>n;
    for(int i=0;i<n;i++)cin>>s[i];

     cout << setprecision(9) << fixed << fun();

}